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655-705 (Hard)|   Algebra|      
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Bunuel
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What is the value of (x^8y^2 + x^2y^8)/(x^2 + y^2)?] 10 Nov 2023, 11:13
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

62% (02:13) correct 38% (02:33) wrong based on 191 sessions

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What is the value of \(\frac{(x^8y^2 + x^2y^8)}{(x^2 + y^2)}\)?

A. \(x^2y^2(x^4-x^2y^2+y^4)\)

B. \(x^2y(x^4+x^2y^2+y^4)\)

C. \(x^2y^2(x^2+y^2)\)

D. \(x^2y^2(x+y)\)

E. \(x^4y^4(x^2-y^2)\)
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A
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VivekSri
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What is the value of (x^8y^2 + x^2y^8)/(x^2 + y^2)?] 10 Nov 2023, 23:10
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Bunuel
What is the value of \(\frac{(x^8y^2 + x^2y^8)}{(x^2 + y^2)}\)?

A. \(x^2y^2(x^4-x^2y^2+y^4)\)

B. \(x^2y(x^4+x^2y^2+y^4)\)

C. \(x^2y^2(x^2+y^2)\)

D. \(x^2y^2(x+y)\)

E. \(x^4y^4(x^2-y^2)\)
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OA is A

Remember whenever these type of questions comes in question which involves algebra with higher power. Try to resolve using putting the value.

Three easy way you can use

1) Put x=y=1
Atleast 2 option will be eliminated for sure
2) Put x=1 and Y =0 or vice versa
If still not resolve
Then put x=1 y=2.
It will mostly resolve your question

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luvsic
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What is the value of (x^8y^2 + x^2y^8)/(x^2 + y^2)?] 27 Jun 2025, 06:32
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numerator = x^2 * y^2 (x^6 + y^6)

x^6 + y^6 can be factored again by sum of cubes
= (x^2 + y^2)(x^4 - x^2 y^2 + y^4)
now the (x^2 + y^2) cancels and you're left with
x^2 y^2 (x^4 - x^2 y^2 + y^4)

A

Bunuel
What is the value of \(\frac{(x^8y^2 + x^2y^8)}{(x^2 + y^2)}\)?

A. \(x^2y^2(x^4-x^2y^2+y^4)\)

B. \(x^2y(x^4+x^2y^2+y^4)\)

C. \(x^2y^2(x^2+y^2)\)

D. \(x^2y^2(x+y)\)

E. \(x^4y^4(x^2-y^2)\)
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What is the value of (x^8y^2 + x^2y^8)/(x^2 + y^2)?] 27 Jun 2025, 06:37
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To solve algebraically, need to know and remember that \((x^3+y^3)=(x+y)(x^2-xy+y^2)\).

So we can rewrite the initial fraction in the following way:

\(\frac{(x^8y^2 + x^2y^8)}{(x^2 + y^2)}\) = \(\frac{(x^2y^2)(x^6+y^6)}{ (x^2 + y^2) }\)

Now we need to factor out \((x^6+y^6)\) = \((x^2)^3+(y^2)^3\) = \( (x^2+y^2)(x^4-x^2y^2+y^4)\)

Finally, we get \(\frac{(x^8y^2 + x^2y^8)}{(x^2 + y^2)}\) = \(\frac{(x^2y^2)(x^6+y^6)}{ (x^2 + y^2) }\)= \(\frac{(x^2y^2) (x^2+y^2)(x^4-x^2y^2+y^4) }{ (x^2 + y^2) }\)

that cancels out \(x^2+y^2\) and we get our answer.
Bunuel
What is the value of \(\frac{(x^8y^2 + x^2y^8)}{(x^2 + y^2)}\)?

A. \(x^2y^2(x^4-x^2y^2+y^4)\)

B. \(x^2y(x^4+x^2y^2+y^4)\)

C. \(x^2y^2(x^2+y^2)\)

D. \(x^2y^2(x+y)\)

E. \(x^4y^4(x^2-y^2)\)
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